Although @znt has gotten the answer through pari, I think it may be instructive to outline my argument.

It all depends on the transition function of Higher Ramification Theory: for a finite extension $K\supset k$ of local fields, $\varphi^K_k$ is a function from $\Bbb R^{\ge0}$ to itself, polygonal and concave, in which, if the (finitely many) vertices are $\{(x_i,y_i)\}$, each $x_i$ is a lower number of a ramification break, and $y_i$ is the corresponding upper number. Perhaps the most useful property of the transition function is that it is functorial: if $L\supset K\supset k$, then $\varphi^L_k=\varphi^K_k\circ\varphi^L_K$.

Although you’ll see below that there are other methods of calculating $\varphi^K_k$, here’s a relatively painless way for totally ramified extensions, *if* you know a prime element $\pi$ of the integers of the big field. Let $F(X)\in k[X]$ be the minimal $k$-polynomial for $\pi$, and draw the Newton copolygon of $F(X+\pi)$.

The copolygon of $g(X)=\sum_na_nx^n$ is the intersection of all the lower halfplanes $\eta\le n\xi+v(a_n)$, where you may take $v$ to be the (additive) valuation for which $v(k^\times)=\Bbb Z$. One sees that the segments of the copolygon are in one-to-one correspondence with the vertices of the Newton polygon: there’s a segment along the line $\eta=n\xi+v(a_n)$ if and only if $(n,v(a_n))$ is a vertex of the polygon.

And you get the transition function from this process by stretching the boundary of the copolygon horizontally by a factor of $e^K_k$, the ramification index (equal to the degree since our extension is presumed totally ramified). That is, every point $(\xi,\eta)$ on the boundary of the copolygon gets moved to the point $(e^K_k\xi,\eta)$ on the graph of the transition function. The function achieved in this way is shifted one unit up and one to the right from that defined in most standard texts, like *Corps Locaux* of Serre. In all cases, the infinite segment in the graph has slope $1/e^K_k$, and in the case of wildly ramified extensions, all slopes are powers of $1/p$.

Now calculate the transition functions of $\Bbb Q_2(3^{1/8})$, $\Bbb Q_2(\sqrt3\,)$, and $\Bbb Q_2(\sqrt2\,)$, all as extensions of $\Bbb Q_2$.
I won’t go through the details, except to say that $(X+1)^8-3$ and $(X+1)^2-3$ are the Eisenstein polynomials for the first two of the fields above. You find that the transition functions are: $\varphi^{\Bbb Q_2(3^{1/8})}_{\Bbb Q_2}$ has vertices $(2,2)$, $(4,3)$, and $(8,4)$;
$\varphi^{\Bbb Q_2(\sqrt3\,)}_{\Bbb Q_2}$ has the single vertex at $(2,2)$, and $\varphi^{\Bbb Q_2(\sqrt3\,)}_{\Bbb Q_2}$ has the single vertex at $(3,3)$.

Even though the transition function of $\Bbb Q_2(\sqrt2,\sqrt3\,)$ over $\Bbb Q_2$ can be calculated directly knowing that a prime element is $\pi=1-(\sqrt3-1)\big/\sqrt2$, you can also get it from the transition functions of the subfields, using functoriality. The upper numbers ($\eta$-values) of the subfields must appear among the upper numbers of the whole extension. Since these numbers are $2$ and $3$, $\varphi^{\Bbb Q_2(\sqrt2,\sqrt3\,)}_{\Bbb Q_2}$ must have the vertices $(2,2)$ and $(4,3)$.

All the above is straightforward and easy. But I wanted the transition function of the extension $\Bbb Q_2(\sqrt2,3^{1/4})$ over $\Bbb Q_2(\sqrt2,\sqrt3\,)$. For that I needed a prime element, and I found
$$
\beta=\frac{\frac{3^{1/4}-1}{\pi}-1}{\pi^2}-1\,,
$$
for which the minimal polynomial over $\Bbb Q_2(\sqrt2,\sqrt3\,)$ is of form $X^2+u_1\pi X+u_2\pi$, for units $u_i$. The upshot is that the transition function has the single vertex $(2,2)$. Compose this function first, then the transition function $\varphi^{\Bbb Q(\sqrt2,\sqrt3\,)}_{\Bbb Q_2}$, and you see that the transition function for the degree-eight extension $\Bbb Q_2(\sqrt2,3^{1/4})$ over $\Bbb Q_2$ has two vertices $(2,2)$ and $(6,3)$. The slope ratio at the lefthand vertex is $4$ rather than $2$.

And that’s enough to get our conclusion that $\Bbb Q_2(\sqrt2,3^{1/8})$ is totally ramified of degree $16$ over $\Bbb Q_2$, since its transition function *must* have a vertex with height $4$, coming from the subfield $\Bbb Q_2(3^{1/8})$. So the full transition function has vertices $(2,2)$, $(6,3)$, and $(14,4)$.